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Gwar [14]
3 years ago
15

What is y=(x+5)(x+4) in standard form?

Mathematics
1 answer:
Helen [10]3 years ago
4 0
It is already in standard form.
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Point A is at the coordinate (-5, 6). Point B is located at (-5, -4). What is the distance between these two points?
katrin [286]

Answer:

10

Step-by-step explanation:

6-(-4)=10

I think so lol

3 0
3 years ago
Read 2 more answers
CONVERT Fahrenheit TO CELSIUS WITH FORMULA
Maksim231197 [3]

The force F1 = 10 N when d1 = 4 m. The square of 4 m is d1² = 16 m². When d = 2 m, its square is d2² = 4 m². Solving for F2

F2× d2² = F1× d1²

F2 × 4m² = 10 N × 16 m²

F2 = (10 N × 16 m^2) / 4 m² F2

= 10 N× (16 m² /4 m²)

F2 = 10 N × 4

F2 = 40 N

The gravitational force produced when they are kept 2 meters apart is 40 N.

5 0
3 years ago
15.
spayn [35]
If I’m correct, I think A. 0.3x2 is the correct answer.
4 0
3 years ago
Read 2 more answers
Monica can walk 3.8 kilometers in 40 minutes, and she can jog twice that distance in the same amount of time. How far could Moni
Elanso [62]
We know that
<span>Monica can walk 3.8 kilometers in 40 minutes
and
</span><span>she can jog twice that distance in the same amount of time
that means
in 40 minutes Monica can jog (3.8 km*2)-----> 7.6 km

1 hour=60 minutes
</span>by proportion<span>
7.6 km/40 minutes=x km/60 minutes
x=7.6*60/40----> x=11.4 km

the answer is
11.4 km</span>
5 0
3 years ago
Help please, I need it
OlgaM077 [116]

Answer:

  2)  c)  (x-3)² + (y+2)² = 25

  5)  x^2 +y^2 -8x -16y +54 = 0

  6)  x^2 +y^2 -10x -12y +36 = 0

Step-by-step explanation:

2) The standard form equation for a circle is ...

  (x -h)^2 +(y -k)^2 = r^2

You are given the center: (h, k) = (3, -2) and a point on the circle. So, the equation will be ...

  (x -3)^2 +(y +2)^2 = r^2

Since we know a point on the circle we know that ...

  (7 -3)^2 +(1 +2)^2 = r^2 = 16 +9 = 25

So, the circle's equation is ...

  (x -3)^2 +(y +2)^2 = 25 . . . . . matches choice C

__

5) As in the previous problem, the standard form equation is ...

  (x -4)^2 +(y -8)^2 = (-1-4)^2 +(7-8)^2 = 25+1 = 26

To put this in general form, we need to subtract 26 and eliminate parentheses.

  x^2 -8x +16 +y^2 -16y +64 -26 = 0

  x^2 +y^2 -8x -16y +54 = 0

__

6) A circle tangent to the y-axis will have a radius equal to the x-value of the center point.

  (x -5)^2 +(y -6)^2 = 5^2

  x^2 -10x +25 +y^2 -12y +36 = 25

  x^2 +y^2 -10x -12y +36 = 0

8 0
4 years ago
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